Wanted to check if I did the implied differentiation correctly:
- Find the tangent line at ($\pi, \pi$) $$ \sin{(x+y)} = 2x -2y$$ $$\cos{(x+y)} \cdot (1+\frac{dy}{dx}) = 2 - (2 \cdot \frac{dy}{dx})$$
$$\cos{(x+y)} + \cos({x+y}) \frac{dy}{dx} = 2 - (2\cdot \frac{dy}{dx})$$ $$\cos{(x+y}) - 2 = -\cos{(x+y)} \frac{dy}{dx} - 2 \cdot \frac{dy}{dx}$$
$$\frac{dy}{dx} (-\cos{(x+y)} + 2) = \cos{(x+y)} - 2$$ $$ \frac{dy}{dx} = -1 \cdot \frac{\cos{(x+y)} - 2}{\cos{x+y} +2}$$
slope at ($\pi, \pi$}: $\frac{- \cos{2\pi} - 2}{\cos{2\pi} + 2} = \frac{1}{3}$
So tangent line: $y - \pi = \frac{1}{3} (x - \pi)$
- Find the tangent line at 1,2
$$y^2 = 5x^4 - x^2$$
$$2y\frac{dy}{dx} = 20x^3 - 2x$$
$$\frac{dy}{dx} = \frac{20x^3 - 2x}{2y}$$
slope at 1,2 = $\frac{20-2}{4}$
tangent line: $y - 2 = \frac{9}{2} \cdot (x-1)$ $$y = \frac{9}{2}x - \frac{5}{2}$$
- I'm confused about how to find the double deriavtive of:
$$x^2 + 4y^2 = 4$$
so
$$y' = 2x +4 \cdot 2y \cdot \frac{dy}{dx} = 0$$ $$2x + 8y \frac{dy}{dx} = 0$$ $$8y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{8y} = \frac{-x}{4y}$$ $$ y'' = \frac{-4y - (-x \cdot 4 \frac{dy}{dx}}{16y^2}$$
$$ \frac{-4y + (4x\frac{dy}{dx})}{16y^2}$$
$$\frac{-4y + \frac{-4x^2}{4y}}{16y^2}$$
1) Here you made a sign mistake
$$\frac{dy}{dx} (-\cos{(x+y)} \color{red}{+ 2}) = \cos{(x+y)} - 2$$
It should be negative not positive... Answer is $$\frac {dy}{dx}=\frac {2-\cos(x+y)}{2+\cos(x+y)}$$
2) $y - 2 = \frac{9}{2} \cdot x(x-1) \to y =\frac{9}{2}x^2 -\frac92x+2$
Maybe you meant $y - 2 = \frac{9}{2} \cdot (x-1) \to y=\frac 92x-\frac 52$
3)$$x^2 + 4y^2 = 4$$ Differentiate and simplify by 2 $$x + 4yy' = 0$$ Differentiate again $$1 + 4((y')^2+yy'') = 0$$ If you want $y''$ as a fucntion of y and x then substitute the value of $y'=-\frac x {4y}$ Substitution $$1 + 4((\frac {x^2} {16y^2})+yy'') = 0$$ Substitute from your first equation the value of $x^2$ $$1 + (\frac {x^2} {4y^2})= -4yy''$$ $$y''=-\frac 1 {4y} - \frac {1-y^2} {4y^3}$$ $$y''=-\frac {1} {4y^3}$$